Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $14$ years; the standard deviation is $1.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $12.7$ years.
Solution: $14$ $12.7$ $15.3$ $11.4$ $16.6$ $10.1$ $17.9$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $14$ years. We know the standard deviation is $1.3$ years, so one standard deviation below the mean is $12.7$ years and one standard deviation above the mean is $15.3$ years. Two standard deviations below the mean is $11.4$ years and two standard deviations above the mean is $16.6$ years. Three standard deviations below the mean is $10.1$ years and three standard deviations above the mean is $17.9$ years. We are interested in the probability of a meerkat living less than $12.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the meerkats will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $12.7$ years and the other half $({16\%})$ will live longer than $15.3$ years. The probability of a particular meerkat living less than $12.7$ years is ${16\%}$.